Can't query database

[DavidPr]

I'm trying to do this:

$query = "SELECT * FROM response WHERE recipe_id='$file' AND allow='1'";
$result = mysql_query($query) or trigger_error("Query: $query\n<br>MySQL Error: " . mysql_error());
if (!$result)
{
echo "";
}
else
{
echo "blah blah";
}

But it's not working. It must be something to do with the AND part but I don't know what. Any clues?

[Dormilich]

mysql_query() returns a resource so I wouldn't check that (this should always return true). do you get any errors?

regards

[Markus]

What datatype is 'allow'. If it's ~int, then you shouldn't have single quotes around it.

Doesn't help when you don't tell use what's isn't happening.

[NeoPa]

David - Check PMs please.

[DavidPr]

"allow" is an INT. The field will either have a 1 or 2 in it. When set to 1 the recipe will be displayed and 2 it will not be displayed.

When a user submits a recipe, the allow field is automatically set to 1. I'm sent an email letting me know that there's a new recipe. I check it to make sure that it is not spam or something and if it's OK I change "allow" to 2, which will make the recipe viewable.

OK, so NO '' around this number. I'll try it and tell you how it works. Thanks.

[Markus]

"allow" is an INT. The field will either have a 1 or 2 in it. When set to 1 the recipe will be displayed and 2 it will not be displayed.

When a user submits a recipe, the allow field is automatically set to 1. I'm sent an email letting me know that there's a new recipe. I check it to make sure that it is not spam or something and if it's OK I change "allow" to 2, which will make the recipe viewable.

OK, so NO '' around this number. I'll try it and tell you how it works. Thanks.

OK, you do that.